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May 6th, 2019, 07:26 AM   #11
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Originally Posted by Integrator View Post

I suppose we have to consider the possibility that $\displaystyle x = a + k(a)i$ where $\displaystyle a \in \mathbb R$ and $\displaystyle i^2=-1$ , which means that the initial equation will result an equation in the form $\displaystyle u + vi=0$ where $\displaystyle u=f(a)$ and $\displaystyle v=g(a)$.How do we solve the equation $\displaystyle f(a)+g(a)i=0$ where $\displaystyle i^2=-1$?Thank you very much!

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Sorry, but I have no idea how one would find complex solutions for the initial equation you provided. Hopefully, someone else with the requisite knowledge can offer you some guidance.
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May 10th, 2019, 09:11 AM   #12
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Hello all,

Solving the WolframAlpha computing program:

The first two solutions given by "WolframAlpha" are the same as those found by me.
From my calculations would also result solutions for $\displaystyle k(a)=0$ , $\displaystyle k(a)=i(a-2)$ , $\displaystyle k(a)=i(a-1)$ and $\displaystyle k(a)=\frac{i(2a-3)}{2}$ where $\displaystyle i^2=-1$ , $\displaystyle a=Re(x)$ and $\displaystyle k(a)=Im(x)$ , but I do not know how to interpret them to calculate $\displaystyle x=a+k(a)i$ ....

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Last edited by Integrator; May 10th, 2019 at 09:34 AM.
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